# Coloring balls with 6 colours

In how many ways can we colour 25 white balls with 6 colours such that

a) each colour is used at least once and no ball remains uncoloured
b) all balls are coloured; I thought it would be $6^{25}$
c) no ball uncoloured, exactly 3 colours used. $\frac{6^{25}}{6\choose 3}$?

• What are your thoughts on (a)? Feb 8, 2018 at 13:59
• If you have distinguishable balls (e.g. if they are numbered or otherwise marked to tell them apart), then your idea for part (b) is correct. But I suspect the balls were meant to be all alike before coloring, so that counting the ways to color them means just different outcomes of six color combinations with repetitions. Feb 8, 2018 at 14:04

I'll wait for your thoughts to respond to (a).

(b) is good: since you have 6 choices for each ball, and 25 balls, the total number of ways is $6 \cdot 6 \cdots 6$ 25 times, or $6^{25}$.

For (c), if you knew a specific set of 3 colors were being used, and it were "up to 3 colors," not "exactly," it would be $3^{25}$, by the same logic as (b). Since we need exactly 3 colors, we need to apply the Principle of Inclusion-Exclusion (are you familiar with this?) to get $3^{25} - 3\cdot2^{25} + 3\cdot1^{25}$. Finally, since there are $\binom63$ choices of sets of 3 colors, the answer should be $\binom63(3^{25} - 3\cdot2^{25} + 3\cdot1^{25})$.

(a) $x_1+x_2+x_3+x_4+x_5+x_6=25$

None of the x values are zero and using STARS and BARS method for positive integer

$\binom{25-1}{6-1} = \binom{24}{5} = \frac{24!}{5!19!} = 42504$

(b) $x_1+x_2+x_3+x_4+x_5+x_6=25$

Using STARS and BARS method for non negative solution

$\binom{25+6-1}{6-1} = \binom{30}{5} = \frac{30!}{5!25!} = 142506$

(c)three colours can be chosen by$\binom{6}{3}$

$x_1+x_2+x_3=25$

Using STARS and BARS method for non negative solution

$\binom{25+3-1}{3-1} = \binom{27}{2} = \frac{27!}{2!25!} = 351$

total=$\binom{6}{3}$*351